Difference between revisions of "HW4-3 Steve Anderson ECE301Fall2008mboutin" - Rhea

Latest revision as of 12:27, 25 September 2008

Part A

CT LTI system:

y(t) = 10x(t-1)

plugging in delta(t) into the system we get:

h(t) = 10delta(t-1)

$ s = j\omega $

$ H(s) = \int_{-\infty}^{\infty}h(t)e^{-st} $

$ H(s) = \int_{-\infty}^{\infty}10delta(t-1)e^{-st} $

$ H(s) = 10\times\int_{-\infty}^{\infty}delta(t-1)e^{-st} $

$ H(s) = 10e^{-s}\, $

Part B

$ x(t) = 4\sin(5 \pi t) - (2 + j)\cos(3 \pi t)\, $

$ y(t) = H(jw)x(t)\, $

$ y(t) = 10e^{-jw}\times[4\sin(5 \pi t) - (2 + j)\cos(3 \pi t)]\, $

$ y(t) = 10e^{-jw}\times [\frac{2}{j}e^{5*j\pi t} - \frac{2}{j}e^{-5*j\pi t} - \frac{2+j}{2}e^{3*j\pi t} - \frac{2+j}{2}e^{-3*j\pi t}]\, $

$ y(t) = \frac{20}{j}e^{5*j\pi t}e^{-jw} - \frac{20}{j}e^{-5*j\pi t}e^{-jw} - (10+5j)e^{3*j\pi t}e^{-jw} - (10+5j)e^{-3*j\pi t}e^{-jw}\, $

$ y(t) = \frac{20}{j}e^{5*j\pi (t-1)} - \frac{20}{j}e^{-5*j\pi (t-1)} - (10+5j)e^{3*j\pi (t-1)} - (10+5j)e^{-3*j\pi (t-1)}\, $